Help with PHYS12 Hw: Beam Hinged at One End, 0.8 kg Rod on Scale

In summary, the conversation is about a student asking for help with their PHYS12 homework and sharing their struggles with understanding their new teacher's Russian accent. They also mention a problem involving a hinged beam and a block, and provide a diagram. The conversation ends with the student thanking for the help and expressing doubts about their teacher's qualifications.
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flash2
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Hey! I'm new here.. and I desperately need help for some PHYS12 hw.. I hope I'm in the right place. My new teacher has a strong russian accent so I don't understand a single word he ever saids.

A uniform 1.3 kg beam hinged at one end supports a 0.5 kg block. the bean is held level by a vertical 0.8 kg rod resting on a Newton scale at the other end. What is reading on the scale?

Here's the diagram:
<img src="http://picturemessaging.fido.ca/mmp...ransform=introtate&value=90&random=12150011">
 
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  • #2
physics_torque.jpg


sry there we go.
 
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We need to see your attempt at solving this problem. The system is in equilibrium, which tells you that the total force and total torgue acting on any object in the system is zero.
 
  • #4
Nvm got it! Pretty much my attempt was totally off. My teacher who's really a sub for our real teacher.. who pretty much ran off to Austrailia gave us teh worksheets and taught the lesson for the worksheets the next class. Anyways my attempt at it was pretty much a blindshot and it dind't hlep that I didn't know what the answer was. Thanks anyways. I'm sure I'll have more questions as my teacher who saids he's a Physics teacher just steals his lesson plans off the internet.

-flash
 

FAQ: Help with PHYS12 Hw: Beam Hinged at One End, 0.8 kg Rod on Scale

How do I calculate the weight of the rod in this scenario?

To calculate the weight of the rod, you will need to use the equation W = mg, where W is the weight, m is the mass, and g is the acceleration due to gravity (9.8 m/s^2). In this case, the mass of the rod is given as 0.8 kg, so the weight would be 0.8 kg x 9.8 m/s^2 = 7.84 N.

What is the equation for the torque on a hinged beam?

The equation for torque on a hinged beam is T = F x d, where T is the torque, F is the force applied, and d is the perpendicular distance from the force to the pivot point. In this scenario, the force is the weight of the rod and the distance is the length of the beam.

How do I find the position of the pivot point on a hinged beam?

To find the position of the pivot point on a hinged beam, you will need to use the equation x = (m1d1 + m2d2) / (m1 + m2), where x is the position of the pivot point, m1 and m2 are the masses on either side of the pivot point, and d1 and d2 are the distances from the pivot point to each mass. In this scenario, m1 would be the mass of the rod and d1 would be the length of the beam.

How do I determine the magnitude and direction of the reaction force at the pivot point?

The magnitude of the reaction force at the pivot point can be found using the equation F = mg + ma, where F is the reaction force, m is the mass of the rod, g is the acceleration due to gravity, and a is the acceleration of the rod (which is 0 in this scenario since it is at rest). The direction of the reaction force is opposite to the direction of the applied force (i.e. the weight of the rod).

Can I use this equation for a beam hinged at both ends?

No, this equation is specifically for a beam hinged at one end. If the beam is hinged at both ends, the equation for torque would be T = F x d, where T is the torque, F is the force applied, and d is the distance from the force to the center of the beam. Additionally, the reaction forces at each end would need to be considered.

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